How do you solve $\frac{1}{x} + 1 \geq 0$ $1/x+1>=0$ using a sign chart?

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How do you solve $\frac{1}{x} + 1 \geq 0$ $1/x+1>=0$ using a sign chart?

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Answer 1 · 2017-09-09T07:20:32

Solution is $x \leq - 1$ $x<=-1$ or $x \geq 0$ $x>=0$ and in interval form it is ${x \in (- \infty, - 1] \cup [0, \infty)}$ ${x in(-oo,-1]uu[0,oo)}$

Explanation:

We can write $\frac{1}{x} + 1 \geq 0$ $1/x+1>=0$ as $\frac{1 + x}{x} \geq 0$ $(1+x)/x>=0$

Hence, for inequality ti satisfy,

we should have sign of both $1 + x$ $1+x$ and $x$ $x$ same i.e.

$1 + x \geq 0$ $1+x>=0$ and $x \geq 0$ $x>=0$ i.e. $x \geq - 1$ $x>=-1$ and $x \geq 0$ $x>=0$ i.e. $x \geq 0$ $x>=0$

or $1 + x \leq 0$ $1+x<=0$ and $x \leq 0$ $x<=0$ i.e. $x \leq - 1$ $x<=-1$ and $x \leq 0$ $x<=0$ i.e. $x \leq - 1$ $x<=-1$

Hence solution is $x \leq - 1$ $x<=-1$ or $x \geq 0$ $x>=0$ and in interval form it is ${x \in (- \infty, - 1] \cup [0, \infty)}$ ${x in(-oo,-1]uu[0,oo)}$

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